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Question Number 35231 by abdo.msup.com last updated on 17/May/18

what is the value of cos(1+i) and cos(1−i)?

$${what}\:{is}\:{the}\:{value}\:{of}\:{cos}\left(\mathrm{1}+{i}\right)\:{and} \\ $$$${cos}\left(\mathrm{1}−{i}\right)? \\ $$

Commented by abdo mathsup 649 cc last updated on 17/May/18

we have the formulaes cos(z)=ch(iz) and sin(z)=sh(iz) ⇒cos(1+i)=ch(i(1+i)) =ch(−1+i) = ((e^(−1+i) +e^(1−i) )/2) =(1/2){ e^(−1) cos(1) +ie^(−1) sin(1) +e cos(1)−iesin(1)} =(1/2){ (e +e^(−1) )cos(1) +i(e^(−1) −e)sin(1)} cos(1−i)=ch(i +1)= ((e^(1+i) +e^(−1−i) )/2) = (1/2){ e cos(1) +ie sin(1) +e^(−1) cos(1)−ie^(−1) sin(1)} =(1/2){ (e +e^(−1) )cos(1) +i( e−e^(−1) )sin(1)}

$${we}\:{have}\:{the}\:{formulaes}\:\:{cos}\left({z}\right)={ch}\left({iz}\right)\:{and} \\ $$$${sin}\left({z}\right)={sh}\left({iz}\right)\:\Rightarrow{cos}\left(\mathrm{1}+{i}\right)={ch}\left({i}\left(\mathrm{1}+{i}\right)\right) \\ $$$$={ch}\left(−\mathrm{1}+{i}\right)\:=\:\frac{{e}^{−\mathrm{1}+{i}} \:\:+{e}^{\mathrm{1}−{i}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\:+{ie}^{−\mathrm{1}} {sin}\left(\mathrm{1}\right)\:+{e}\:{cos}\left(\mathrm{1}\right)−{iesin}\left(\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left({e}\:+{e}^{−\mathrm{1}} \right){cos}\left(\mathrm{1}\right)\:+{i}\left({e}^{−\mathrm{1}} \:−{e}\right){sin}\left(\mathrm{1}\right)\right\} \\ $$$${cos}\left(\mathrm{1}−{i}\right)={ch}\left({i}\:+\mathrm{1}\right)=\:\frac{{e}^{\mathrm{1}+{i}} \:+{e}^{−\mathrm{1}−{i}} }{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:{e}\:{cos}\left(\mathrm{1}\right)\:+{ie}\:{sin}\left(\mathrm{1}\right)\:+{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)−{ie}^{−\mathrm{1}} {sin}\left(\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left({e}\:+{e}^{−\mathrm{1}} \right){cos}\left(\mathrm{1}\right)\:+{i}\left(\:{e}−{e}^{−\mathrm{1}} \right){sin}\left(\mathrm{1}\right)\right\} \\ $$

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